t2+47t=0

Solution for t2+47t=0 equation:

t2+47t=0

We add all the numbers together, and all the variables

t^2+47t=0

a = 1; b = 47; c = 0;
Δ = b2-4ac
Δ = 472-4·1·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-47}{2*1}=\frac{-94}{2}
=-47
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+47}{2*1}=\frac{0}{2}
=0
$